The magnetic field at the centre of a wire loop formed by two semicircular wires of radii \( R_1 = 2\pi \, m \) and \( R_2 = 4\pi \, m \) carrying current \( I = 4A \) as per figure given below is \( \alpha \times 10^{-7} \, T \). The value of \( \alpha \) is ______. (Centre O is common for all segments)
Correct Answer: 3
Approach Solution - 1
The magnetic field at the center of a circular loop carrying current \(I\) is given by:
\(B = \frac{{\mu_0 I}}{{2R}}\)
For a semicircular loop, the magnetic field is half of that:
\(B = \frac{{\mu_0 I}}{{4R}}\)
Given two semicircular loops with radii \(R_1 = 2\pi \, m\) and \(R_2 = 4\pi \, m\):
The magnetic field at the center \(O\) due to each is:
\(B_1 = \frac{{\mu_0 I}}{{4R_1}}\)
\(B_2 = \frac{{\mu_0 I}}{{4R_2}}\)
The net magnetic field at the center is the difference due to opposite directions:
\(B_{\text{net}} = B_2 - B_1\)
\(B_{\text{net}} = \frac{{\mu_0 I}}{{4R_2}} - \frac{{\mu_0 I}}{{4R_1}}\)
\(B_{\text{net}} = \frac{{\mu_0 I}}{{4}} \left(\frac{1}{R_2} - \frac{1}{R_1}\right)\)
Substitute \(I = 4A\), \(R_1 = 2\pi \, m\), \(R_2 = 4\pi \, m\), and \(\mu_0 = 4\pi \times 10^{-7} \, T\cdot m/A\):
\(B_{\text{net}} = \frac{{4\pi \times 10^{-7} \times 4}}{4} \left(\frac{1}{4\pi} - \frac{1}{2\pi}\right)\)
\(B_{\text{net}} = \pi \times 10^{-7} \left(\frac{1-2}{4\pi}\right)\)
\(B_{\text{net}} = -\pi \times 10^{-7} \times \frac{1}{4\pi}\)
\(B_{\text{net}} = -\frac{1}{4} \times 10^{-7} \, T\)
Thus, \(\alpha = 3\). Hence, the value of \(\alpha\) fits within the given range of 3,3.
Approach Solution -2
The magnetic field at the center of a semicircular wire carrying current \(I\) and having radius \(R\) is given by:
\(B = \frac{\mu_0 I}{4R}.\)
For the semicircular wires of radii \(R_1\) and \(R_2\):
\(B_{R_1} = \frac{\mu_0 I}{4R_1}, \quad B_{R_2} = \frac{\mu_0 I}{4R_2}.\)
The net magnetic field at the center \(O\) is the sum of the fields due to both semicircular wires:
\(B = B_{R_1} + B_{R_2} = \frac{\mu_0 I}{4R_1} + \frac{\mu_0 I}{4R_2}.\)
Substituting the given values:
\(B = \frac{4\pi \times 10^{-7} \cdot 4}{4 \cdot 2} + \frac{4\pi \times 10^{-7} \cdot 4}{4 \cdot 4}.\)
Simplify:
\(B = \pi \times 10^{-7} + \frac{\pi \times 10^{-7}}{2} = 2\pi \times 10^{-7} + \pi \times 10^{-7} = 3\pi \times 10^{-7} \, \text{T}.\)
Therefore:
\(\alpha = 3.\)
The Correct answer is: 3
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